以毫秒为单位计算 C 程序中的经过时间 [英] Calculating elapsed time in a C program in milliseconds

问题描述

我想计算执行我的程序的某些部分所花费的时间(以毫秒为单位).我一直在网上寻找，但关于这个主题的信息并不多.你们有人知道怎么做吗?

I want to calculate the time in milliseconds taken by the execution of some part of my program. I've been looking online, but there's not much info on this topic. Any of you know how to do this?

推荐答案

最好的回答方式是举例:

Best way to answer is with an example:

#include <sys/time.h>
#include <stdlib.h>
#include <stdio.h>
#include <math.h>

/* Return 1 if the difference is negative, otherwise 0.  */
int timeval_subtract(struct timeval *result, struct timeval *t2, struct timeval *t1)
{
    long int diff = (t2->tv_usec + 1000000 * t2->tv_sec) - (t1->tv_usec + 1000000 * t1->tv_sec);
    result->tv_sec = diff / 1000000;
    result->tv_usec = diff % 1000000;

    return (diff<0);
}

void timeval_print(struct timeval *tv)
{
    char buffer[30];
    time_t curtime;

    printf("%ld.%06ld", tv->tv_sec, tv->tv_usec);
    curtime = tv->tv_sec;
    strftime(buffer, 30, "%m-%d-%Y  %T", localtime(&curtime));
    printf(" = %s.%06ld
", buffer, tv->tv_usec);
}

int main()
{
    struct timeval tvBegin, tvEnd, tvDiff;

    // begin
    gettimeofday(&tvBegin, NULL);
    timeval_print(&tvBegin);

    // lengthy operation
    int i,j;
    for(i=0;i<999999L;++i) {
        j=sqrt(i);
    }

    //end
    gettimeofday(&tvEnd, NULL);
    timeval_print(&tvEnd);

    // diff
    timeval_subtract(&tvDiff, &tvEnd, &tvBegin);
    printf("%ld.%06ld
", tvDiff.tv_sec, tvDiff.tv_usec);

    return 0;
}

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